$T$ is a linear transformation in $\mathbb{R}^3$ that rotates vectors 30 degrees counterclockwise about the vector $\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$.
I need to first, come up with a basis $C$ that can be used to get $[T]_C^C$ without performing any computations, and second, find $[T]_C^C$.
I was thinking of using a basis such as $C=\left\{\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\right\}$. Therefore, we could just rotate vectors about the "second" vector in the basis, effectively making the transformation matrix $[T]_C^C$ look like a rotation about the y-axis in the standard basis of $\mathbb{R}^3$.
Am I on the right track with this? I just don't know if the first and third vectors matter and was also considering using $\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$ in place of $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ but I was thinking that it might not matter.
How would I then go about constructing $[T]_C^C$ in each case? Would any basis with $\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ in the second position yield the same result?
I'm thinking that using either of the bases I mentioned, $[T]_C^C$ would come out to something like $\begin{pmatrix} \cos{30} & 0 & -\sin{30} \\ 0 & 1 & 0 \\ \sin{30} & 0 & \cos{30} \end{pmatrix}$ or $\begin{pmatrix} \cos{30} & 0 & \sin{30} \\ 0 & 1 & 0 \\ -\sin{30} & 0 & \cos{30} \end{pmatrix}$, but I'm really not sure how to figure any of this out without computations. What am I missing?